 Solution

First reduce the unicyclic graph to its unique graph cycle $C$. Delete the edges of $C$ but retain every vertex. The remaining graph is a forest, since any surviving graph cycle would be a second one in $G$. Each of its connected components of a graph contains a vertex of $C$: follow a graph path to $C$ in the original connected graph, stopping at the first vertex of $C$; this initial segment uses no edge of $C$.

No such connected component of a graph contains two vertices of $C$. Otherwise, on the tree graph path between them, choose two successive visits to $C$. The segment between these visits has no internal vertex on $C$ and uses no edge of $C$. Combining it with either connecting arc of $C$ gives a second graph cycle, a contradiction. Thus the remaining graph consists of trees, each rooted at exactly one vertex of $C$.

After choosing a proper vertex colouring of $C$ from $q$ colours, colour these rooted trees away from their roots. Each of the $n-\ell$ off-graph cycle vertices has one already coloured parent in a rooted tree, leaving exactly $q-1$ choices. The multiplication principle therefore givesEquivalently, this repeatedly applies the chromatic polynomial after attaching a leaf formula.

To compute the chromatic polynomial of a cycle, the base case is the graph triangle:For $s\ge 4$, deleting a cycle graph edge gives a path graph on $s$ vertices, while edge contraction gives $C_{s-1}$. The chromatic polynomial of that path graph is $x(x-1{)}^{s-1}$, by choosing the colour of its first vertex and then each successive vertex. Hence the deletion-contraction recurrence for the chromatic polynomial givesUsing mathematical induction, substitute $p_{C_{s-1}}(x)=(x-1{)}^{s-1}+(-1{)}^{s-1}(x-1)$ to obtainMultiplying by the factor for the attached trees proves the chromatic polynomial of a unicyclic graph formula